Let $$\overrightarrow a = 2\overrightarrow i + \overrightarrow j + \overrightarrow k ,\,\overrightarrow b = \overrightarrow i + 2\overrightarrow j - \overrightarrow k $$        and a unit vector $$\overrightarrow c $$ be coplanar. If $$\overrightarrow c $$ is perpendicular to $$\overrightarrow a $$ then $$\overrightarrow c = ?$$  

Solution :
$$\eqalign{ & \overrightarrow c = t\overrightarrow a + s\overrightarrow b .{\text{ Also }}\left| {\overrightarrow c } \right| = 1\left( {{\text{given}}} \right) \cr & {\text{Now, }}\overrightarrow c = t\left( {2\overrightarrow i + \overrightarrow j + \overrightarrow k } \right) + s\left( {\overrightarrow i + 2\overrightarrow j - \overrightarrow k } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {2t + s} \right)\overrightarrow i + \left( {t + 2s} \right)\overrightarrow j + \left( {t - s} \right)\overrightarrow k \cr & \therefore \,\,\left| {\overrightarrow c } \right| = 1\,\,\,\therefore \,{\left( {2t + s} \right)^2} + {\left( {t + 2s} \right)^2} + {\left( {t - s} \right)^2} = 1.....\left( 1 \right) \cr & {\text{and }}\overrightarrow a .\overrightarrow c = 0\,\, \Rightarrow 2\left( {2t + s} \right) + 1\left( {t + 2s} \right) + 1\left( {t - s} \right) = 0 \cr & {\text{or }}6t + 3s = 0{\text{ or }}2t + s = 0 \cr & \therefore \,{\text{from}}\left( 1 \right),\,{0^2} + {\left( {t - 4t} \right)^2} + {\left( {t + 2t} \right)^2} = 1 \cr & {\text{or }}18{t^2} = 1\,\,\,\,\,\therefore t = \pm \frac{1}{{3\sqrt 2 }} \cr & \therefore \,s = - 2t = \mp \frac{{\sqrt 2 }}{3} \cr & \therefore \,\overrightarrow c = \left( {2t + s} \right)\overrightarrow i + \left( {t + 2s} \right)\overrightarrow j + \left( {t - s} \right)\overrightarrow k \cr & = 0\overrightarrow i - 3t\overrightarrow j + 3t\overrightarrow k \cr & = - 3t\left( {\overrightarrow j - \overrightarrow k } \right) \cr & = - 3.\frac{{ \pm 1}}{{3\sqrt 2 }}\left( {\overrightarrow j - \overrightarrow k } \right) \cr} $$